## Thomas' Calculus 13th Edition

a) $3x+5y+4z=18$ and b) $x=3+6t,y=5+10t; z=-4+8t$
a. As we know that the vector equation is given by:$r(x,y,z)=r_0+t \nabla f(r_0)$ The equation of tangent line is given as: $\nabla f(3,5,-4)=\lt 6,10,8 \gt$ Thus, $6(x-3)+10(y-5)+8(z+4)=0$ or, $6x+10y+8z=6 \implies 3x+5y+4z=18$ b. As we know that the vector equation is given by:$r(x,y,z)=r_0+t \nabla f(r_0)$ The parametric equations for $\nabla f(3,5,-4)=\lt 6,10,8 \gt$ are: $x=3+6t,y=5+10t; z=-4+8t$